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Q.
If A.M. of two numbers is twice of their G.M., then the ratio of greatest number to smallest number is
Sequences and Series
Solution:
Let the numbers be $x$ and $y$.
Since, $ \frac{x+y}{2}=2 \sqrt{x y} $
$ \Rightarrow x+y =4 \sqrt{x y} ....$(i)
$ \Rightarrow (x+y)^2 =16 x y $
Also, $ (x-y)^2 =(x+y)^2-4 x y $
$\therefore (x-y)^2 =16 x y-4 x y=12 x y $
$ \Rightarrow x-y =2 \sqrt{3 x y}.....$(ii)
On solving Eqs. (i) and (ii), we get and
$x=(2+\sqrt{3}) \sqrt{x y}$
and $ y=(2-\sqrt{3}) \sqrt{x y}$
$\therefore $ Required ratio $ =\frac{x}{y}=\frac{(2+\sqrt{3}) \sqrt{x y}}{(2-\sqrt{3}) \sqrt{x y}} $
$ =(2+\sqrt{3})^2=7+4 \sqrt{3}$